A compound has the following % composition. Zn = 22.65%; S = 11.15%; O = 61.32% and H = 4.88%. It’s relative molecular mass is 287 g. Calculate it’s molecular formula assuming that all the hydrogen in the compound is present in combination with oxygen as water of crystallization. [Zn = 65, S = 32, O = 16, H = 1]

Question

A compound has the following % composition. Zn = 22.65%; S = 11.15%; O = 61.32% and H = 4.88%. It&#8217;s relative molecular mass is 287 g. Calculate it&#8217;s molecular formula assuming that all the hydrogen in the compound is present in combination with oxygen as water of crystallization. [Zn = 65, S = 32, O = 16, H = 1]

Flash Institute · Accepted Answer

Element
% composition
At. wt.
Relative no. of atoms
Simplest ratio

Zn
22.65
65
22.65/65​ = 0.3484
0.3484/0.3484​ = 1

S
11.15
32
11.15/32​ = 0.3484
0.3484/0.3484​ = 1

O
61.32
16
61.32/16​ = 3.832
3.832/0.3484 = 11

H
4.88
1
4.88/1​ = 4.88
4.88/0.3484​ = 14

Simplest ratio of whole numbers = Zn : S : O : H
= 1 : 1 : 11 : 14
Hence, empirical formula is ZnSO11H14
Molecular weight = 287
Empirical formula weight = 65 + 32 + 11(16) + 14(1)
= 65 + 32 + 176 + 14
= 287
n = Molecular\ weight\over Emperical\ Formula\ Weight
= 287\over 287 = 1
Molecular formula = (ZnSO11H14)1 = ZnSO11H14
Since all the hydrogen in the compound is in combination with oxygen as water of crystallization.
Therefore, 14 atoms of hydrogen and 7 atoms of oxygen = 7H2O and hence, 4 atoms of oxygen remain.
Hence, the molecular formula is ZnSO4.7H2O.

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Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Zn	22.65	65	= 0.3484	= 1
S	11.15	32	= 0.3484	= 1
O	61.32	16	= 3.832	= 11
H	4.88	1	= 4.88	= 14

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